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IRINA_888 [86]
3 years ago
7

I have four dogs whose average weight is 63 pounds. I also have three cats. The average weight of all seven of my animals is 41

pounds. What is the average weight of my cats?
Mathematics
1 answer:
Brums [2.3K]3 years ago
8 0

Answer: 11.67 pounds

Step-by-step explanation:

From the question, the person have four dogs whose average weight is 63 pounds. Then, the total weight of the dogs will be:

= 63 pounds × 4

= 252 pounds

There are also three cats and the average weight of all seven of my animals is 41 pounds. This means that the total animals have a title weight of:

= 41 pounds × 7

= 287 pounds.

To get the weight of the cats, we subtract the total dog's weight from that of the whole animals. This will be:

= 287 - 252

= 35 pounds.

Since there are 3 cats, their average weight will be:

= 35/3

= 11.67 pounds

The average weight of the cats is 11.67 pounds.

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What is the first step to isolate the variable term on one side of the equation?
Vilka [71]

Answer:

the answer to the whole equation is x=30.

Step-by-step explanation:

do 2/3(2) and -1/2(3)

4/6x and -3/6x

add the -3/6x to the other side

7/6x or 1 and 1/6x=5

times both by 6

x=30

5 0
3 years ago
A researcher has developed a new drug designed to reduce blood pressure. In an experiment, 21 subjects were assigned randomly to
Paladinen [302]

Answer:

Step-by-step explanation:

Hello!

The objective of the research is to compare the newly designed drug to reduce blood pressure with the standard drug to test if the new one is more effective.

Two randomly selected groups of subjects where determined, one took the standard drug (1- Control) and the second one took the new drug (2-New)

1. Control

X₁: Reduction of the blood pressure of a subject that took the standard drug.

n₁= 23

X[bar]= 18.52

S= 7.15

2. New

X₂: Reduction of the blood pressure of a subject that took the newly designed drug.

n₂= 21

X[bar]₂= 23.48

S₂= 8.01

The parameter of study is the difference between the two population means (no order is specified, I'll use New-Standard) μ₂ - μ₁

Assuming both variables have a normal distribution, there are two options to estimate the difference between the two means using a 95% CI.

1) The population variances are unknown and equal:

[(X[bar]₂-X[bar]₁)±t_{n_1+n_2-2;1-\alpha /2}*(Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  })]

t_{n_1+n_2-2;1-\alpha /2}= t_{23+21-2;1-0.025}= t_{42;0.975}= 2.018

Sa=\sqrt{\frac{(n_1-1)*S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{22*7.15^2+20*8.01^2}{42} }= 7.57

[23.48-18.52]±2.018*(7.57*\sqrt{\frac{1}{21} +\frac{1}{23}  })]

[0.349; 9.571]

2) The population variables are unknown and different:

Welche's approximation:

[(X[bar]₂-X[bar]₁)±t_{Dfw;1-\alpha /2}*( \sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} })]

Df_{w}= \frac{(\frac{S_1^2}{n_1} +\frac{S^2_2}{n_2} )^2}{\frac{(\frac{S_1^2}{n_1} )^2}{n_1-1}+ \frac{(\frac{S_2^2}{n_2} )^2}{n_2-1}  } =  \frac{(\frac{7.15^2}{23} +\frac{8.01^2_2}{21} )^2}{\frac{(\frac{7.15^2}{21} )^2}{20}+ \frac{(\frac{8.01^2}{23} )^2}{22}  } = 42.85= 42

t_{Df_w;1-\alpha /2}= t_{42; 0.975}=  2.018

[(23.48-18.52)±2.018\sqrt{\frac{7.15^2}{23} +\frac{8.01^2}{21} }]

[0.324; 9.596]

I hope this helps!

4 0
3 years ago
Describe and correct the error in determining wheather (8,11) is a solution of y-x=3
Burka [1]

(8, 11) is the correct solution. x = 8, y = 11

Plug it in:

11-8=3

That's correct.

7 0
3 years ago
Read 2 more answers
4px = 20 <br> (find p) <br> -4px = 20<br> (find x)
Kazeer [188]

Answer:

4px=20

x=(20/4p)-----(1)

-4px=20--------(2)

putting value of x in epn 2

-4p*20=20*4p

-80p=+80p

Both cut these value are rejected

Checkout the question again bro it might be wrong

Please for my hard work make me brainliest answer

5 0
3 years ago
In the system shown below, what are the coordinates of the solution that lies in quadrant I?
Arisa [49]

Answer:

(4,3)

Step-by-step explanation:

Given

x^2+y^2=25

x-y^2=-5

In order to solve the equations, from equation 2 we get

-y^2= -5-x

y^2=5+x

Putting the value of y^2 in equation 1

x^2+5+x=25

x^2+5-25+x=0

x^2+x-20=0

x^2+5x-4x-20= 0

x(x+5)-4(x+5)=0

(x+5)(x-4)=0

So

x+5=0  x-4=0

x=-5   x=4

Now for x=-5

x^2+y^2=25

(-5)^2+y^2=25

25+y^2=25

y^2=25-25

y^2=0

so Y=0  

And for x = 4

x^2+y^2=25

(4)^2+y^2=25

16+y^2=25

y^2=25-16

y^2=9

y= ±3

So the solution to the system of equations is

(-5,0) , (4,3), (4,-3)

The only solution that belongs to first quadrant is (4,3)

4 0
3 years ago
Read 2 more answers
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